Rog Delta Filter - Infn-lnf

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Resonant Bars: Time dispersion and efficiency detection for impulsive signals of known shape using the “ROG Delta Filter” Sabrina D’Antonio for the ROG Collaboration INFN sez. University of Rome “Tor Vergata “ Archana Pai, Pia Astone INFN Sez. University of Roma “La Sapienza” Status of the EXPLORER- NAUTILUS resonant bars EXPLORER 2005: Nautilus 20-05-2005 S n ( f ) ≈ 10 −21 / Hz Explorer:5-05-2005 BW ~ 50 Hz at 10 −20 / T=3 K Teff ~ 2mK Hz NAUTILUS 2005: S n ( f ) ≈ 1.5 *10 −21 / Hz Sqrt(T/MQ) BW ~ 40 Hz at 10 −20 / T=3.3K Teff ~ 2mK Hz Investigating the Data Analysis Issues for Improved Bandwidth (BW) • Presently the resonant bars Explorer and Nautilus are working with broader BW. • It is worth investigating different kind of signals in addition to the delta signals as done in the past. • It is necessary to reassess the assumption of deltalike signals for matched filtering as the filtered output crucially depends on the shape of the waveform. • In this study we discuss this problem with an example of GW signals: quasi-normal-modes of proto-NS during the cooling phase after the supernova core collapses: Damped sinusoid evolving in frequency and damping time. Investigating the Data Analysis Issues for Improved Bandwidth (BW) Topic of this talk: Validate ROG delta filters : a). Loss in SNR using ROG delta filters compared to SNR for Matched filter (in case of known signal) b). Accuracy in the signal arrival time measured by ROG delta filter compared to that obtained with the Matched filter. Important while searching coincidences between resonant bars c) Amplitude signal estimation using the ROG delta filter (Ad) compared to that obtained with the matched filter (Am). signal parameters: [?] where : we can use the Delta matched filter we have to use a “bank of filters” Investigating the Data Analysis Issues for Improved Bandwidth (BW) First Step (discussed in this presentation): In this preliminary work we consider damped sinusoid signals of known shape with constant frequency Fo and damping time τ . This study is performed by analytical approach & Simulations: injection of signals of known shape at the input of adaptive delta/signals matched filters We have used Explorer data “Dec 2004”: Resonant frequencies at the minus/plus mode=Fm=904.7Hz/Fp=927.45Hz. Second Step: (Future Work) Study for signals of unknown shape Evolving f-mode with A=10^-20 Evolving g-mode with A=10^-20 Recent simulations [Shibata&Uryu,2002 Simulation] have shown that a merger of two NS with equal mass and low compactness can give rise to short lived supra massive NS. Typical energy during merger ∆E ≈ 10 M c −2 2 sun The QNM in such a merger may have more energy than those during the cooling phase after the core collapse ∆E ≈ 10 M . c −4 2 sun Layout-ROG Detector + Filters U(t) A Uo(t) h(t) Bartransducer ADC Read-out Matched Filter Gm Signal :Damped Sinusoid with constant fo,τ: h(t ) = e − (t −to ) / τ sin(ωo (t − to )) *θ (t − to ) Force 2 2 x l L is the effective lenght L = 4 π mx is the reduced mass 2 m = M /2 Response of Bar-Transducer to h(t) : x SNRd (delta matched filter)=A‹Uo,Gδ›/σd SNRm (matched filter)=A/ σm σd, σm :std dev of the filtered noise L U ( jω ) = W ( jω ) * F ( jω ) = −ω m *W ( jω ) * H ( jω ) 2 2 ux x A Amplitude, A ~ A(τ,Fo,ho,S) S=System parameters : L ∂ h(t ) F (t ) = m 2 ∂t A δ Delta Filter Gδ ux System transfer function Response of the bar-transducer to the damped sinusoid (summary) τ u(t ) = A+ (t,τ , Fo , S ) *sin(ω+t + φ+ ) − A− (t,τ , Fo, S ) *sin(ω−t + φ− ) − Ao (t,τ , Fo ) *sin(ωot + φo ) Case I:Fo=Fm,τ :[10 ms - 200 ms] 10 ms Maximum amplitude A=u(To) vs To increases with τ τ τ 20 ms 50 ms 200 ms τ= 200 ms For Fo=Fm : For τ ≥ 50 ms A+< 1 /(π * BW ) ,τg increases with For ! (Independent of detectors' bandwidth) Max(g(t)=g(To)) is not wel-peaked. 100 ms With noise, the error in To can be large ! 200 ms τg[s] Decay time [s] of g(t) vs τ [s] For Explorer τ > 1 /(π * BW ) 0.8 s 1s τ Assuming the Bw ~ 1KHz τ. MATCHED FILTER SNR Explorer:Dec 2004 (Data used for simulations) For a detectable SNR=5, min(ho) vs τ fo=915 fo=900 fo=fm=927.45 fo=930 SNR vs τ for damped sine signal with ho =10^-19 10^-19 fo=fm=904.7 For a detectable SNR=5, min(hrss) vs τ 0.35 fo=fm=904.7 fo=915 fo=fp=927.41 fo=900 fo=930 fo=900 fo=fm=904.7 fo=915 0.05 fo=930 fo=fp=927.41 Damped sinusoid filtered with Delta Filter Case fo=fm, τ [10 ms-0.6 s] τ=10 ms τ=0.1s τ=20 ms τ=0.2s τ=30 ms τ=0.4s τ=50 ms τ=0.6 s τ=20 ms Delta Filter: SNR COMPARISON SNR-Delta Filter to SNR-Matched Filter Explorer:Dec 2004 (Data used for simulations) I- fo:[840,890] 840 840 880 880 885 @ τ* ~ 25 ms 915 890 BW sig.~ 40 Hz II- fo:[900,915] 915 905 Fm=904.7 910 900 I- Fo far from the resonant frequencies, for τ <τ* the loss in the SNR decreases with t because the frequency band of the signal falls in the more sensitive band of the detector till t=t* -> important to apply the matched filter for τ<τ*. For τ >τ* the loss in the SNR increases with τ : the signal stays in the detector for more time =>important to apply the matched filter II- Fo near the resonant frequencies The loss in the SNR rapidly increases with τ Delta Filter: SNR COMPARISON SNR-Delta Filter to SNR-Matched Filter Fo:[840,890] Hz < Fm = 904.7 Hz 840 880 885 1. Fo< Fm (Fo > Fp) and away from the freq -band SNR_D ~ SNR_M, 890 τ 30 ms Fo:[900,915] Hz around Fm=904.7 Hz 915 900 τ 20 ms 904.7 Accepted loss in SNR with Delta filter ~ 10% 905 910 2. For [900, 930], Max(τ) to be searched Fo Fm=904.7 910 915 Max(τ) 18ms 20ms 12ms Q ~51 ~57 ~35 DELTA FILTER : Time Dispersion and Amplitude Estimation Time dispersion :∆t Vs τ Ad/Am: Vs τ 880 Hz τ 927.4 904.7 880 950 τ 915 1. Time dispersion : For Fo near the two resonant frequencies ∆t increases with τ. Constant after τ >1sec is misleading Important while fixing the coincidence window For Fs away from the freq-band, ∆t ->0 msec 915 Hz 904.7 Hz 2.The error in the amplitude estimation is For Fo near the two resonant frequencies the error increases faster with τ For Fs away from the freq-band. The error is ~ 20% Conclusions on the use of the Delta Filter Signal parameters Fo<880 Hz Fo>935Hz Dt Error in the amplitude estimation 0 15% τ <100ms Fo near Fm,Fp τ <20 ms Loss in SNR 10% Delta filter is ok Increases with < Few ms τ Increases with <30% τ Increases with <10% Delta filter is ok Bank of Matched filters should be used for the other cases ! τ Delta Filter: SNR COMPARISON SNR-Delta Filter to SNR-Matched Filter fo:[840,890] fo:[925,930] 840 880 885 890 fo:[900,915] fo:[935,950] Response of the bar-transducer ... fo=860 Fo=850 Tau = 0.1,0.2,0.4,0.6 ms Response of the bar-transducer to the damped sinusoide Case III :Fm < Fo < Fp , let τ=30 ms F0=905 Hz F0=920 Hz F0=910 Hz F0=925 Hz F0=915 Hz Response of the bar-transducer to the damped sinusoide (summary) Case II :Fm < Fo < Fp , τ increasing from 30 ms till 1 s τ=30 ms τ=800 ms τ=50 ms τ=1 s τ=100 ms τ=200 ms The maximum amplitude A=u(to) is a function of τ to is constant in this particular case: Fo=(Fm+Fp)/2 Response of the bar-transducer to the damped sinusoide Case I:Fo=Fm, τ increasing from 10 ms till 1 s ms τ=10 τ= 800 ms τ= 20 ms τ=100 ms τ= 200 ms τ=1 s The maximum amplitude A=u(to) is a linear function of τ. Layout-ROG Detector + Filters U(t) A δ Delta Filter Gδ A Uo(t) h(t) Bartransducer ADC Read-out Matched Filter Gm Amplitude, A ~ A(tau,Fo,ho,S) S=System parameters ROG Delta Filter:Gδ=NdW*ux(f)/(S(f)*Md) SNR=A‹Uo,Gδ›/Nd Matched filter:Gm=NmUo*(f)/(S(f)) SNR=A/Nm S(f) :Noise power spectra at the output 1 | U ( jω ) | =[ ∫ dω ] 2π S (ω ) Nm,Nd :std dev of the filtered noise 2 N 2 o m A 1 | W ( jω ) | N =[ ∫ dω ] π 2 S (ω )( M ) 2 −1 2 ux 2 d d −1 Layout-ROG Detector + Filters U(t) A Uo(t) h(t) Bartransducer Read-out ADC Amplitude, A ~ A(tau,Fo,ho,S) A δ Delta Filter Gδ Matched Filter Gm S=System parameters ROG Delta Filter:Gδ=NdW*ux(f)/(S(f)*Md) SNR=A‹Uo,Gδ›/Nd Matched filter:Gm=NmUo*(f)/(S(f)) SNR=A/Nm S(f) :Noise power spectra at the output A Nm,Nd :std dev of the filtered noise Minimum energy (Mc^2) to be emitted in QNM to obtain SNR=5 930 915 900 904.7 Response of the bar-transducer to the damped sinusoid (summary) Case II : Fo = (Fp +Fm)/2 = 915, τ : [10 ms till 200ms] τ=100 ms ~44 ms τ=10 ms τ=200 ms τ=20 ms τ τ=50 ms 1.Maximum amplitude A=u(To) increases with τ 2. To is constant. Depends on the modulation freq of the envelope i.e. (Fp – Fm) ~ 22.7Hz Damped sinusoid filtered with Delta Filter CASE fo=(fm+fp)/2, τ [10 ms- 1s] τ=10 ms τ=0.8s τ=20 ms τ=1s τ=0.1s τ=0.2s τ=0.2s The temporal position of Max(A ) independent of τ. Because the time To corre. to max(u(t)) is independent of τ